[vc_row][vc_column][vc_column_text]Let t be the (positive) square root of 2. What is the 62nd digit after the decimal point of (1+t) to the power 2012?
Even if your calculator were able to help with this, you don't need it!
Hint: It would be nice to work with an expression without the awkward square root of 2.[/vc_column_text][/vc_column][/vc_row][vc_row][vc_column][lvca_accordion][lvca_panel panel_title="Reveal Solution"]Let t be the positive square root of 2.
We can expand (1+t)^2012 by the Binomial Theorem to get
1 + 2012t + (2012.2011t^2)/2 + ... + t^2012 (1)
and since even powers of t are powers of 2, all the terms with even powers of t are integers. Can we get rid of the odd powers?
Consider (1‐t)^2012. This is
1 ‐ 2012t + (2012.2011t^2)/2 ‐ ... + t^2012 (2)
where the coefficients are the same as in (1) except that the terms with odd powers of t have opposite signs. So if we add (1) and (2), we get
(1+t)^2012 + (1‐t)^2012 = an integer (3)
since the only terms which contribute all have even powers of t and are therefore integers.
Now |1‐t| < 1/2, and since 2^10 > 10^3, (1/2)^2012 < 10^(‐670) which means that the first 670 digits after the decimal point of (1‐t)^2012 are all zeroes. So (3) tells us that the first 670 digits of (1+t) after the decimal point are all 9s, and therefore the 62nd is a 9.
[/lvca_panel][/lvca_accordion][/vc_column][/vc_row]
